Lesson Video: Kinetic Energy | Nagwa Lesson Video: Kinetic Energy | Nagwa

Lesson Video: Kinetic Energy

In this video, we will learn how to calculate the kinetic energy of a moving particle of mass 𝑚 that moves with a velocity of magnitude 𝑣.


Video Transcript

In this video, we’re going to learn about kinetic energy. We’ll learn what this term means, how to calculate it, and how it relates to the work done on an object.

To start out, imagine that you’re at the local bowling alley playing a few games. After bowling several frames with a regular bowling ball, you decide to try an experiment. What if, you wonder, instead of using a bowling ball, you use a basketball and are able to throw it faster towards the pins? The basketball is clearly lighter than a bowling ball. But you think the extra speed you’ll be able to give it will make up for that loss of mass.

To understand how to compare bowling with a bowling ball versus with a basketball, we’ll want to know something about kinetic energy. There are many different types of energy. Kinetic energy is a type having to do with motion. Any mass that’s in motion has kinetic energy. As an equation, we often symbolize kinetic energy with capital KE. And it’s equal to one-half an object’s mass times its speed squared.

Let’s notice a few things about this equation. First, we see that kinetic energy is a scalar quantity. It has a magnitude, but no direction associated with it. The 𝑣 in this equation is a speed, that is, not a velocity vector. Another thing we notice is that kinetic energy depends linearly on mass. This means that, for a given speed, if the mass of our object is doubled, our kinetic energy of that object is doubled too.

On the other hand, we have a speed squared term in this equation, which means that speed is the dominant driver of an object’s kinetic energy. If the speed of an object doubles, then the kinetic energy of that object goes up by a factor of four. Kinetic energy, the energy due to motion, is related to the work done on an object. When we think about the work done on an object, we recall that it’s the force multiplied by the distance the object travels.

Looking at that equation, we see that force, by Newton’s second law of motion, is equal to an object’s mass times its acceleration. After making that substitution, we see that we can make another one for the displacement, 𝑑. Looking over the list of kinematic equations, we see that the second equation involves this displacement, 𝑑. And if we rearrange this expression to solve for 𝑑, we can take what 𝑑 equals and substitute it in for 𝑑 into our expression for work. And when we do this, we see that the acceleration cancels out of the numerator and denominator.

This simplifies our expression to work is equal to one-half mass times the final speed squared minus the initial speed squared. Recalling the expression we just learned for kinetic energy, we can see that this expression for work involves the difference between a final kinetic energy and an original kinetic energy. So work equals final kinetic energy minus initial kinetic energy. Or using the Greek symbol Δ to represent change, it equals ΔKE.

This expression has a name. It’s called the work-energy theorem. It simply says that if you measure the amount of work done on an object, that will equal the change in the kinetic energy of that object. Both work and kinetic energy then are expressed in units of joules. Let’s get some practice working with kinetic energy and the work-energy theorem through a couple of examples.

An automobile of mass 2.00 times 10 to the third kilograms has a velocity of 1.00 times 10 to the two kilometers per hour. A runner of mass 80 kilograms has a velocity of 10.0 meters per second. And an electron has a velocity of 2.0 times 10 to the seventh meters per second. What is the kinetic energy of the automobile? What is the kinetic energy of the runner? What is the kinetic energy of the electron?

For each of these three objects — the automobile, the runner, and the electron — we want to solve for the kinetic energy. We’ll call these values KE sub 𝑎, KE sub 𝑟, and KE sub 𝑒, respectively. And we’ll use these subscripts of 𝑎, 𝑟, and 𝑒 to record the masses and speeds given in this problem statement.

We’re told the respective masses and speeds for all three of these objects, except for the mass of the electron, which we can look up. We’ll approximate that mass as exactly 9.1 times 10 to the negative 31st kilograms. When we recall that an object’s kinetic energy is equal to one-half its mass times its speed squared, we see that, to solve for the kinetic energy of the car, the runner, and the electron, we’ll need to have those two pieces of information. And indeed, we do have that information for each of these three objects. So we’re ready to calculate their kinetic energies.

The kinetic energy of the automobile is equal to one-half its mass times its speed squared. When we plug these values in, looking at the speed value, we were given the car’s speed in units of kilometers per hour. But we know if we divide that speed magnitude by a factor of 3.6, we’ll convert it to units of meters per second, consistent with the units in the rest of our expression. Having done that unit conversion, we’re ready to calculate KE sub 𝑎. When we do, we find it’s 772 kilojoules or 772000 joules. That’s the automobile’s kinetic energy.

Next, we’ll calculate the runner’s kinetic energy. That’s equal to half the runner’s mass multiplied by the runner’s speed squared. Plugging these values in, we see that they’re already in SI base units and we’re ready to calculate. When we do, we find that the runner’s kinetic energy is 4.00 kilojoules. That’s their energy of motion at this speed.

Lastly, we wanna calculate the electron’s kinetic energy, one-half its mass times its speed squared. Plugging in the electron’s mass and the speed at which it’s moving, when we calculate this value, we find it’s equal to 1.8 times 10 to the negative 16th joules. That’s the kinetic energy of the electron.

Now let’s look at an example involving the work-energy theorem.

A child is pulling two red wagons, with the second one tied to the first by a nonstretching rope. Each wagon has a mass of 5.0 kilograms. If the child exerts a force of 40 newtons for 6.0 meters, how much has the kinetic energy of the two-wagon system changed?

In this statement, we’re told the mass of each wagon, 5.0 kilograms, which we can write down as 𝑚. We’re also told that the child pulls on the wagon system with a force of 40 newtons, which we’ll call 𝐹, and that this pull happens for a distance of 6.0 meters, which we’ll label 𝑑. We want to solve for the change in the kinetic energy of the two-wagon system. We’ll call this ΔKE.

To get started on our solution, we can recall that kinetic energy and work done on an object are related. The work-energy theorem tells us that work equals change in kinetic energy for a given object. In our case, we can write that ΔKE equals 𝑤 since we wanna solve for ΔKE. And we can use the fact that work is also equal to the force exerted on an object multiplied by the distance it travels, so long as that force and the distance traveled are in the same direction.

Using that fact, we could write that ΔKE equals 𝐹 times 𝑑, where 𝐹 and 𝑑 are given to us. Plugging these two values in, when we calculate ΔKE, we find it’s equal to 240 joules. That’s the work done on this two-cart system. And by the work-energy theorem, it’s equal to the change in kinetic energy of the system.

Let’s summarize what we’ve learned so far about kinetic energy. We’ve seen that kinetic energy is energy due to motion. And it’s equal to one-half an object’s mass times its speed squared. We’ve also seen that kinetic energy is related to work through the work-energy theorem — that the work done on an object is equal to its change in the kinetic energy. And lastly, we’ve seen that kinetic energy is measured in units of joules. Kinetic energy, the energy of motion, is one of the two major types of energy.

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